2017 AIME II Problems/Problem 10
Problem
Rectangle has side lengths and . Point is the midpoint of , point is the trisection point of closer to , and point is the intersection of and . Point lies on the quadrilateral , and bisects the area of . Find the area of .
Solution 1
Impose a coordinate system on the diagram where point is the origin. Therefore , , , and . Because is a midpoint and is a trisection point, and . The equation for line is and the equation for line is , so their intersection, point , is . Using the shoelace formula on quadrilateral , or drawing diagonal and using , we find that its area is . Therefore the area of triangle is . Using , we get . Simplifying, we get . This means that the x-coordinate of . Since P lies on , you can solve and get that the y-coordinate of is . Therefore the area of is .
Solution 2 (No Coordinates)
Since the problem tells us that segment bisects the area of quadrilateral , let us compute the area of by subtracting the areas of and from rectangle .
To do this, drop altitude onto side and draw a segment parallel to \overline{AD}\overline{ND}M\overline{AD}\overline{OE}a\triangle{MOQ}~\triangle{COD}\triangle{DOC}\triangle{AND}BCONP\overline{MC}\triangle{BPC}\overline{PF}b\overline{BC}\overline{PG}\overline{DC}\overline{PF}=\overline{GC}\triangle{MDC}\triangle{PGC}\triangle{CDP}$ is given by ~blitzkrieg21 and jdong2006
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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